Prove that $x^2 + y^2 \equiv 3\,mod\,m$ has and integer solution for every m

Question

I am struggling with this proof: Prove that $x^2 + y^2\equiv 3\,mod\,m$ has and integer solution for all m. I have tried: setting x or y to zero. Can't do that because then the solution wouldn't be integer. I have also tried to make $x^2 + y^2 = 3$, which would give me the proof for all m, but again I am unable to get an integer solution. Any help is appreciated!

Answer 1

This is not true for all integers $m$. For example, consider $m = 4$. Since squares are congruent to $0$ or $1$ modulo $4$, the sum of $2$ squares can only be congruent to $0$, $1$ or $2$ modulo $4$, so there's no solution for $x$ and $y$ in

$$x^2 + y^2 \equiv 3 \pmod 4 \tag{1}\label{eq1A}$$